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In the Given Figure, Two Circles Touch Each Other Externally at Point P. Ab is the Direct Common Tangent of These Circles. Prove That: (Ii) Angles Apb = 90° - Mathematics

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Question

In the given figure, two circles touch each other externally at point P. AB is the direct common tangent of these circles. Prove that: 

(ii) angles APB = 90°

 

Solution

ii) Now in Δ ATP ,
∴ `∠`TAP = `∠`TPA
Similarly in Δ BTP,`∠`TBP = `∠`TPB
Adding,
`∠`TAP +`∠`TBP =`∠`APB
But
∴ TAP + `∠`TBP  + `∠`APB =180°
⇒ `∠`APB =  `∠`TAP  + `∠`TBP =90°

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APPEARS IN

 Selina Solution for Concise Mathematics for Class 10 ICSE (2020 (Latest))
Chapter 18: Tangents and Intersecting Chords
Exercise 18 (A) | Q: 13.2 | Page no. 275
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In the Given Figure, Two Circles Touch Each Other Externally at Point P. Ab is the Direct Common Tangent of These Circles. Prove That: (Ii) Angles Apb = 90° Concept: Tangent Properties - If Two Circles Touch, the Point of Contact Lies on the Straight Line Joining Their Centers.
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